The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is
The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is
Sum of digits
1 + 2 + 3 + 5 + 6 + 7 = 24
So, either 3 or 6 rejected at a time
Case 1 Last digit is 2
……….2
no. of cases = 
Case 2 Last digit is 6
……….6
= 4! = 24
Total cases = 72
Similar Questions for you
Start with
(1)
(2)
(3) GTE : 4!, GTN: 4!, GTT : 4!
(4) GTWENTY = 1
⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553
x + 2y + 3z = 42
0 x + 2y = 42 ->22 cases
1 x + 2y = 39 ->19 cases
2 x + 2y = 36 ->19 cases
3 x + 2y = 33 ->17 cases
4 x + 2y = 30 ->16 cases
5 x + 2y = 27 ->14 cases
6 x + 2y = 24
Total ways to partition 5 into 4 parts are:
5 0
4 1 0
3 2 0
3 1 0
2 1
51 Total way
After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
15+3–1C2 = 17C2 ways
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Maths NCERT Exemplar Solutions Class 11th Chapter Seven 2025
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